SOLUTION: Consider the equation f(x)=x^2+6x+4 Find the x-intercepts by completing the square. Find the vertex Find the axis of symmetry I tried working out this problem 3 times and i c

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Question 308363: Consider the equation f(x)=x^2+6x+4
Find the x-intercepts by completing the square.
Find the vertex
Find the axis of symmetry
I tried working out this problem 3 times and i can not solve it. My teacher said that when i find the x-intercepts that i should leave my answer in square roots rather than the decimal approximations.
Found 2 solutions by scott8148, CharlesG2:
Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website!
x-intercepts are when f(x)=0

0 = x^2 + 6x + 4

c term should be 9 for a perfect square (half of the b coefficient, squared)

adding 5 ___ 5 = x^2 + 6x + 9 = (x + 3)^2

taking square root ___ �√5 = x + 3 ___ -3�√5 = x

axis of symmetry ___ x = -b / (2a) ___ x = -3

vertex is on the axis of symmetry ___ f(-3) = (-3)^2 + 6(-3) + 4 = -5

vertex ___ (-3, -5)

Answer by CharlesG2(834) (Show Source):
You can put this solution on YOUR website!
"Consider the equation f(x)=x^2+6x+4
Find the x-intercepts by completing the square.
Find the vertex
Find the axis of symmetry
I tried working out this problem 3 times and i can not solve it. My teacher said that when i find the x-intercepts that i should leave my answer in square roots rather than the decimal approximations."
f(x) = x^2 + 6x + 4
set f(x) = 0 and complete the square
coefficient of 6x is 6, half of 6 is 3, and 3 squared is 9,
and 9 - 5 = 4
0 = (x + 3)^2 - 5
5 = (x + 3)^2
sqrt(5) = x + 3 OR -sqrt(5) = x + 3
-3 + sqrt(5) = x OR -3 - sqrt(5) = x (these are your x-intercepts)
parabolic equation is in the form f(x) = a(x - h)^2 + k, where (h,k) are the coordinates of the vertex
a = 1 and is positive so parabola opens upward
h = -3 and k = - 5 so vertex is at (-3, -5)
and axis of symmetry is x = h or x = -3